Atomic Nucleus Assignment Help

Atomic Nucleus

It is an important experiment, which led Rutherford to the discovery of atomic nucleus.

An alpha particle is helium nucleus containing 2 protons and neutrons. Therefore, an alpha particle has 4 units of mass and two units of positive charge. Many radioactive elements emit alpha particles. The experimental set up used by Rutherford and his collaborators, Geiger and Marsden is shown in fig.1 and 2.

S is a speck of a radioactive source contained in a lead cavity. The alpha particles emitted by the source are collimated into a narrow beam with the help of a lead slit (collimator). The collimated beam is allowed to fall on a thin gold foil of thickness of the order of 2.1 × 10-7 m. the α particles scattered in different directions are observed through a rotatable detector consisting of a zinc sulphide screen and a microscope. The alpha particles produce bright flashes or scintillations on the ZnS screen. These are observed in the microscope and counted at different angles from the direction of incidence of the beam. The angle θ of deviation of an alpha particle from its original direction is called its scattering angle θ.

Observations: a graph is plotted between the scattering angle θ and the number of α particles N (θ), scattered at ∠ θ for a very large number of α particles, shown in fig.

The dots in this figure represent the data points of the actual experiment. The solid curve is the theoretical prediction based on the assumption that atom has a small, dense, positively charged nucleus. We find that

(i) Most of the alpha particles pass straight through the gold foil. It means they do not suffer any collision with gold atoms.

(ii) Only about 0.14% of incident α particles scatter by more than 1°.

(iii) About one α particle in every 8000α particles deflects by more than 90°.

Explanation: the scattering of α particles is due to Coulombian interaction of α particles with positive charges and electrons in every atom of the gold foil. If positive charges and electrons were distributed uniformally in an atom, (as per Thomson model), the scattering angle θ would be very small.

An α particle is over 7000 times more massive than an electron, and in this experiment, α particle is travelling at a high speed, therefore, very strong forces alone could have deflected them through large angles.

This led Rutherford to postulate that the entire positive charge of the atom must be concentrated in a tiny central core of the atom. This tiny central core of each atom was called atomic nucleus.

As the gold foil is very thin, it can be assumed that α particles will suffer not more than one scattering during their passage through it. An α particle carries two units of positive charge and has mass of helium atom. Charge on gold nucleus + Ze, where atomic number of gold, Z = 79. As gold nucleus is about 50 times heavier than an α particle, we assume that it would remain stationary in the scattering process. Therefore, the trajectory of α particle can be computed using Newton’s second law of motion and Coulomb force of repulsion betweenα particle and gold nucleus, i.e.

F = [1/4∏ε0 × (Ze) (2e)]/r2

Where r is distance of α particle from the centre of the nucleus, the magnitude and direction of the force of an α particle changes continuously as it approaches the nucleus first and then recedes away from it.

As shown in figure, an alpha particle (1), tending to collide head on with the nucleus, slows down due to repulsive force of the nucleus, finally stops and is then repelled back. This α particle, therefore, retraces its path, scattering through 180°.

Alpha particles2, 2’ tending to hit the nucleus at its periphery, experience strong repulsive forces and get scattered through large angles (θ > 90°)

The alpha particles 3, 3’ which pass at a distance from the nucleus, experience small repulsive forces and get scattered through small angles. The α particles which pass at still larger distances from the nucleus go almost undeviated.